Edit: Thanks for showing me in the answers that above formula fails if $A$ and $B$ are disjoint but their boundaries still intersect. I was able to come up with a similar formula which avoids this case
$$[\partial(A\cap B)]\setminus(\partial A \cap \partial B)= (A \cap \partial B) \cup (\partial A \cap B),$$
which I was able to prove and suffices for what I need to do.

However, when showing that $ (A \cap \partial B) \cup (\partial A \cap B)\subseteq \partial(A\cap B)$, I needed to assume that the topology is induced by a metric. I wonder if the formula still holds in an arbitrary topological space.

$\begingroup$As usual with equalities between sets, the most straight-forward way is to check that each set is a subset of the other. There might be other, cleverer methods out there, but this one, at least, will not fail.$\endgroup$
– ArthurNov 26 '16 at 16:34

3 Answers
3

It does not hold. Consider for example $$A=\{x\in \mathbb{R}^n:|x|<1\}$$ and
$$B=\{x\in \mathbb{R}^n:|x-(2,0,\dots,0)|<1\}.$$
Since $A\cap B=\varnothing$, $\partial(A\cap B)=\varnothing$, but the RHS in your formula is the set $\{(1,0,\dots,0)\}$.

$\begingroup$Your example uses tow disjoint sets whose boundaries still intersect. It seems that this is only case when the formula does not hold. Please check the updated question.$\endgroup$
– J.DoeNov 27 '16 at 11:56